If a+2b=6 then find Maxima of ab and a²b. Pleaseget me solution I am eager to know this

this should be done by differentiating seperately i mean partial differention the other way is that to sub no other way than this the best way is partial differentiation

so, a=b for max. value

3a=6, a=2=b

So max value of ab=3 x 3=9

     and that of a²b=3² x 3=27

sorry there is mistake in above answer. So here is the correct explanation for this

We can apply AM>or=GM here

so for finding max value of ab

(a+2b)/2>or=(2ab)^1/2

3²>or=2ab

ab

and for finding max value of a²b

we have write a+2b=a/2+a/2+2b

now (a/2+a/2+2b)/3>or=(a²b/2)^1/3

2³>or=a²b/2

so a²b⁴ so max value of a²b is 2⁴

 

sorry there is mistake in above answer. So here is the correct explanation for this

We can apply AM>or=GM here

so for finding max value of ab

(a+2b)/2>or=(2ab)^1/2

3²>or=2ab

ab

and for finding max value of a²b

we have write a+2b=a/2+a/2+2b

now (a/2+a/2+2b)/3>or=(a²b/2)^1/3

2³>or=a²b/2

 

a +2b = 6 

a = 6-2b

Now, 

        ab= 6b-2b²

For maxima and minima,

         (d/db){6b-2b²} = 0

                6-4b = 0

then,        b = 3/2

                d²/db²{6b-2b²} = -4     …....for all vlaues of d

             Hence, ab is maximum for b = 3/2 

             ab = 9/2 for b = 3/2

             Similarly, we can get a²b = 16.